Let the plane P: 8 x+_1 y+_2 z+12=0 be parallel to the line L: (x+2)/(2)=(y-3)/(3)=(z+4)/(…

Question

Let the plane $\mathrm{P}: 8 x+\alpha_{1} y+\alpha_{2} z+12=0$ be parallel to

the line $\mathrm{L}: \frac{x+2}{2}=\frac{y-3}{3}=\frac{z+4}{5}$. If the

intercept of $\mathrm{P}$

on the $y$-axis is 1 , then the distance between $\mathrm{P}$ and $\mathrm{L}$ is :

Accepted Answer

P: $8 x+\alpha_{1} \mathrm{y}+\alpha_{2} \mathrm{z}+12=0$

L: $\frac{\mathrm{x}+2}{2}=\frac{\mathrm{y}-3}{3}=\frac{\mathrm{z}+4}{5}$

$\because \mathrm{P}$ is parallel to $\mathrm{L}$

$\Rightarrow 8(2)+\alpha_{1}(3)+5\left(\alpha_{2}\right)=0$

$\Rightarrow 3 \alpha_{1}+5\left(\alpha_{2}\right)=-16$

Also $y$-intercept of plane $P$ is 1

$\Rightarrow \alpha_{1}=-12$

And $\alpha_{2}=4$

$\Rightarrow$ Equation of plane $\mathrm{P}$ is $2 \mathrm{x}-3 \mathrm{y}+\mathrm{z}+3=0$

$\Rightarrow$ Distance of line L from Plane $\mathrm{P}$ is

$=\left|\frac{0-3(6)+1+3}{\sqrt{4+9+1}}\right|$

$=\sqrt{14}$

Solutions · 1